A 100 W sodium lamp is radiating light of wavelength `5890A`, uniformly in all directions,
a. At what rate, photons are emitted from the lamp?
b. At what distance from the lamp, the average flux is 1 photon(`cm^2-s)^-1?`
c. What are the photon flux and photon density at 2m from the lamp?

Explanation

The energy of photon is given by
`E=hupsilon=(hc)/(lamda)=(1990xx10^(-28)J)/(lamda)`
`=(1990xx10^(-28))/(5890A)=3376xx10^(-22)J` ..(i)
Given that the lamp is emitting energy at the rate of `100Js^(-1)` (power=100 W). Hence, number of photons N emitted is given by
`N=(100)/(3376xx10^(-22))cong3xx10^(20)` photons `s^(-1)` ..(ii)
b. We regard the lamp as a point source. Therefore, at a distance r from the lamp, the light energy is uniformly distributed over the surface of sphere of radius r. So, N photons are crossing area `4pir^2` of spherical surface per second. So, flux at a distance r is given by
`n=(N)/(4pir^2)` or `r=sqrt(((N)/(4pi)))`
or `r=sqrt(((3xx10^(20))/4xx3.14))cm=488860km` ...(iii)
So, at this distance, on the average one photon will cross through `1 cm^(2)` area normal to tradial direction.
c. Photon flux at `t=2m`
`n=sqrt(((3xx10^(20))/(4xx3.14)))cm=5.9xx10^(14)` photons `cm^(-2)`
Average density of photons at `r=2m `is given by
`rho=(3xx10^(20))/(4xx3.14xx(200)^(2)xx(3xx10^(10)))`
`=2xx10^(4)` photons `cm^(-2)`